A solution containing a metal ion \(\left(\mathrm{M}^{3+}(a q)\right)\) is electrolyzed by a current of \(5.0 \mathrm{~A}\). After \(10.0\) minutes, \(1.19 \mathrm{~g}\) of the metal is plated out. (a) How many coulombs are supplied by the battery? (b) What is the metal? (Assume \(100 \%\) efficiency.)

To calculate the charge supplied by the battery, we need to use the formula for electrical charge: \(Q = It\) where \(Q\) is the charge (in coulombs), \(I\) is the current (in amperes), and \(t\) is the time (in seconds). We are given the current \(I = 5.0 \,\text{A}\) and the time \(t = 10.0 \,\text{min}\). To convert minutes into seconds, we multiply the time by 60: \(t = 10.0 \,\text{min} \times 60 = 600 \,\text{s}\). Now, we can calculate the charge: \(Q = (5.0 \,\text{A})(600 \,\text{s}) = 3000 \,\text{C}\). The battery supplies 3000 coulombs of charge. #b) Identify the metal involved in the plating process#

To identify the metal, we can use Faraday's second law of electrolysis, which states that the mass of a substance produced at an electrode is proportional to its molar equivalent weight and the charge passed through the electrolyte: \(m = \dfrac{M}{nF}Q\) where \(m\) is the mass of the substance produced (in grams), \(M\) is the molar mass of the substance (in \(\text{g/mol}\)), \(n\) is the number of electrons exchanged per formula unit, \(F\) is the Faraday's constant (96500 \(\text{C/mol}\) of electron), and \(Q\) is the charge passed (in coulombs). We are given that the mass \(m = 1.19 \,\text{g}\), the metal ion has a charge of \(3+\), meaning \(n = 3\). Also, we calculated the charge passed, \(Q = 3000 \,\text{C}\). Rearranging the formula above to find the molar mass \(M\): \(M = \dfrac{mnF}{Q} = \dfrac{(1.19 \,\text{g})(3)(96500 \,\text{C/mol})}{(3000 \,\text{C})} = 74.97 \,\text{g/mol}\) Looking up the periodic table, the metal with a molar mass close to 74.97 \(\text{g/mol}\) is iron (Fe), with a molar mass of 55.85 \(\text{g/mol}\) and 3 electrons exchanged per formula unit when forming the \(\text{Fe}^{3+}\) ion. Therefore, the metal involved in the plating process is iron.